[tex]x' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x - vt) [/tex]

[tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(t - \frac{vx}{c^2}) [/tex]

So I think I am starting to understand x'. When the observer O looks at the point x, where an object or event would be, The distance from O' to that object or event is contracted. Lets say that the observer O looks out and sees that O' is 5 meters from an object at x. This length will have to be scaled by gamma in the O' frame because O is actually seeing a a contracted length that appears to be 5 meters. the uncontracted length that o' observes will be 5m*gamma which is what the formula for x' says.

However time is giving me more difficulties. It appears that the formula is saying that the time interval (t') that has elapsed for O' is longer by a factor of gamma than the interval that O sees has elapsed (t) on his watch.

This seems opposite to how I usually think of time dilation which is that time for a moving observer will pass more slowly. When a spaceship flies by, the clock on board appears to only make a half turn while ours on earth makes a full turn. To me, the Lorentz transformation seems to be saying that more time will elapse on the spaceship than for us, or that our clock will only make a half turn while theirs make a full turn. What in my thinking is wrong?

Thanks for the assistance.